Problem 108 - Elementary and Intermediate Algebra

Question

Perform the operations. Write all answers in the form \(a+b i .\) $$ \frac{3}{2 i^{17}} $$

Step-by-Step Solution

Answer

The expression simplifies to \(0 - \frac{3}{2}i\).

Step 1: Simplify the Exponent

First, simplify the expression involving the imaginary unit. Since the powers of \(i\) cycle every four (\(i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1\)), find the equivalent power of \(i\) for \(i^{17}\). Calculate the remainder of 17 when divided by 4: \(17 \div 4 = 4 \text{ R } 1\). Thus, \(i^{17} = i^1 = i\).

Step 2: Substitute and Simplify

Replace \(i^{17}\) with \(i\) in the original expression: \(\frac{3}{2i}\). To simplify the division by \(i\), multiply both the numerator and the denominator by \(-i\) to eliminate the imaginary unit from the denominator: \[ \frac{3}{2i} \times \frac{-i}{-i} = \frac{3(-i)}{2(i)(-i)} = \frac{-3i}{2(-i^2)} \].

Step 3: Simplify the Result

Since \(i^2 = -1\), we have \(-i^2 = 1\). Thus, the expression becomes: \[ \frac{-3i}{2 \times 1} = \frac{-3i}{2} \]. Therefore, the expression in the form \(a + bi\) is \(0 - \frac{3}{2}i\).

Key Concepts

Imaginary Unit

The imaginary unit, represented as \(i\), is a cornerstone of complex numbers. Its definition is simple yet fascinating: it is the square root of -1. Since no real number squared equals a negative number, we use \(i\) to represent \(\sqrt{-1}\). In doing so, \(i\) unlocks an entirely new number system called complex numbers.
These complex numbers consist of two parts: a real part and an imaginary part. For example, in the complex number \(a + bi\), \(a\) is the real part, and \(bi\) is the imaginary part. The imaginary unit \(i\) allows us to explore mathematical operations beyond those confined to just the real number line.
\(i\) is crucial for equations that arise in engineering, physics, and various applied mathematics disciplines. Being comfortable with \(i\) helps students expand their understanding of mathematics beyond the typical scope of real numbers.

Powers of i

Understanding the powers of \(i\) is vital for simplifying expressions involving imaginary numbers. The powers of \(i\) recycle every four steps, forming a predictable cycle:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

This cycle repeats for any higher powers. For instance, \(i^5\) is the same as \(i^{1}\), \(-i\), and so on. To determine any power of \(i\), you can simply divide the exponent by 4 and observe the remainder:

• If the remainder is 1: \(i^n = i\)
• If the remainder is 2: \(i^n = -1\)
• If the remainder is 3: \(i^n = -i\)
• If the remainder is 0: \(i^n = 1\)

In our example, \(i^{17}\) has a remainder of 1 when divided by 4, thus it equals \(i^1 = i\). Comprehending this cycle helps streamline computations involving complex numbers.

Simplifying Expressions with i

When simplifying expressions with \(i\), especially when \(i\) appears in the denominator, the goal is to rationalize it. This means we try to eliminate the \(i\) from the denominator by using multiplicative methods. Consider how we simplify \(\frac{3}{2i}\) from the exercise:
We can multiply the numerator and the denominator by \(-i\) to remove \(i\) from the denominator:
\[\frac{3}{2i} \times \frac{-i}{-i} = \frac{3(-i)}{2(i)(-i)} = \frac{-3i}{2(-i^2)}\]Next, recall that \(i^2 = -1\). Therefore, \(-i^2 = 1\). Now our expression simplifies further to:
\[\frac{-3i}{2 \times 1} = \frac{-3i}{2}\]
By this method, the complex expression is simplified to the standard form \(a + bi\), in this instance, \(0 - \frac{3}{2}i\). Rationalizing denominators with \(i\) through complex conjugates or direct multiplication ensures the expression is in its cleanest form. This practice is essential for clarity and consistency in mathematical communication.